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solns5_600

# solns5_600 - ECE-600 Homework#5 Introduction to Digital...

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ECE-600 Introduction to Digital Signal Processing Winter 2011 Homework #5 Feb. 18, 2011 HOMEWORK SOLUTIONS #5 1. (a) Linear convolution y [ n ] = k = −∞ w [ k ] x [ n k ] can be understood as flipping one of the sequences and “sliding” it through the other sequence, where “sliding” involves pointwise multiplication and summing over the result: 1 2 3 4 result n = 0 : 3 2 1 1 n = 1 : 3 2 1 2 + 2 = 4 n = 2 : 3 2 1 3 + 4 + 3 = 10 n = 3 : 3 2 1 4 + 6 + 6 = 16 n = 4 : 3 2 1 8 + 9 = 17 n = 5 : 3 2 1 12 Meanwhile, in Matlab, we get >> conv([1,2,3],[1,2,3,4]) ans = 1 4 10 16 17 12 (b) N -circular convolution can be written in two ways: y [ n ] = N summationdisplay k =0 w [ k ] x [ ( n k ) N ] = summationdisplay k = −∞ w [ ( k ) N ] x [ n k ] , The first can be viewed as flipping and sliding cyclic sequences, while the second can be viewed as periodically extending one sequence and then doing a linear convolution. We illustrate the latter approach here (where N = 4): ··· 1 2 3 4 1 2 3 4 1 2 3 4 ··· result n = 0 : 3 2 1 1 + 8 + 9 = 18 n = 1 : 3 2 1 2 + 2 + 12 = 16 n = 2 : 3 2 1 3 + 4 + 3 = 10 n = 3 : 3 2 1 4 + 6 + 6 = 16 Noting that the output sequence is periodic with period N = 4, we only need to compute these 4 outputs. (c) Recalling the DFT convolution property: N 1 summationdisplay m =0 x [ m ] w [ ( n m ) N ] N -DFT ←→ X [ k ] W [ k ] we know that pointwise multiplication in the N -DFT domain equals N -circular convolution in the time domain. Thus, starting with two time domain signals, we can accomplish circular convolution by i) taking the N -DFT of each sequence, ii) pointwise multiplying the DFT outputs, and iii) taking the N -IDFT of the result. Matlab yields: >> ifft(fft([1,2,3],4).*fft([1,2,3,4],4))

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